Here's a classic puzzle (I think Martin Gardner talks about it somewhere, though I'm not sure exactly where):
Alice and Bob are co-conspirators. Alice is dealt five random cards from a standard deck; she looks at the cards, and then shows four of them to Bob in some chosen order. Using only the four cards and the order they are shown in, Bob must determine the identity of the fifth card.
The solution that I know of is:
Alice and Bob agree on an ordering of the $52$ cards, and a bijection between $S_3$ to $\{1,\dots,6\}$. With this information, an ordering of any three cards gives an integer between $1$ and $6$ inclusive. Now, Alice uses the pigeonhole principle to find two cards of the same suit. Without loss of generality, their ranks are $r_1$ and $r_2$ where $r_1+a \equiv r_2 \pmod{13}$ and $1 \leq a \leq 6$ (if they weren't originally, swap $r_1$ and $r_2$). Alice puts the card of rank $r_1$ first (naming the suit) and then orders the other three cards in a way that tells Bob $a$ (naming the rank).
Notice that $52$ cards is the maximum possible number that this solution works for, since it depends heavily on there being only four suits and thirteen cards in each suit. However, it seems like it wastes a fair amount of information whenever
the suit distribution isn't 2-1-1-1.
Is there a method to do this which works on a deck containing more than $52$ cards? The obvious upper bound on deck size is that Bob sees a list of four ordered cards and can deduce from it the original five cards Alice was dealt, so it it works for a deck of size $n$ we must have $$ 4!\binom{n}{4} \geq \binom {n}{5} $$ or $n \leq 124$. But that still leaves quite a gap...
